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2 years ago

6

A student sees a newspaper ad for an apartment that has 1330 square feet (ft^2) or floor space. How many square meters of area a

re there?

1 answer:

Natasha2012 [34]2 years ago

6 0

<span>=1330 ft^2 x (1m^2/3.25 ft^2)
=124 m^2

</span>

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A compound whose molecules contain one boron atom and three fluorine atoms would be named monoboron fluoride. Please select the

Slav-nsk [51]

I know it is false...............

5 0

3 years ago

4. The atmosphere is composed of about 78% nitrogen, 21% oxygen, and 1% argon. Typical atmospheric pressure in Boulder, Colorado

photoshop1234 [79]

<u>ANS</u>

<u>Step</u> <u>1</u> :Explanation of required formula.

According to Dalton's Law of Partial Pressures, the partial pressure of a component of a gaseous mixture depends on the mole ratio of said component and the total pressure of the gaseous mixture

i.e Pi=Xi × Ptotal,

Here we don't know exactly how many moles of the mixture we have,

but we know that 78.0% of all the molecules present in the mixture are nitrogen molecules, 21.0% are oxygen molecules, and 1% are molecules of Ar gas.

As we know, a mole is simply a very large collection of molecules. In order to have one mole of a substance, we need to have 6.022 × 1023 molecules of that substance.

This means that the actual number of moles is not important here, because the ratio that exists between the number of molecules is equivalent to the ratio that exists between the number of moles.

Hence,

<u>Step</u> <u>2</u> : Calculate of mole fraction of the mixture.

mole fraction of nitrogen = 78 /100 = 0.78

mole fraction of O2 =

21 /100 = 0.21

mole fraction of Argon =

1 /100 = 0.01.

<u>Step</u> <u>3</u> : Calculate the pressure contributed by each of the mixture.

The pressure contributed by N2 = mole fraction of N2 × Total pressure = 0.78 × 0.83 atm = 0.6474 atm

The pressure contributed by O2 = 0.21 × 0.83 atm = 0.1743 atm

The pressure contributed by N2 = 0.01 × 0.83 atm = 0.0083 atm.

<u>Tha</u><u>nk</u> <u>You</u> !!!!!!

6 0

2 years ago

A new planet has been discovered and given the name Planet X . The mass of Planet X is estimated to be one-half that of Earth, a

harina [27]

Answer:

vₐ = v_c $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

Em_f = U = - G Mm / R₂

energy is conserved

Em₀ = Em_f

½ m v_c² - G Mm / R = - G Mm / R₂

v_c² = 2 G M (1 /R - 1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

v_c = $\sqrt{\frac{2GM}{R} }$

now indicates that the mass and radius of the planet changes slightly

M ’= M + ΔM = M ( $1+ \frac{\Delta M}{M}$ )

R ’= R + ΔR = R ( $1 + \frac{\Delta R}{R}$ )

we substitute

vₐ = $\sqrt{\frac{2GM}{R} } \ \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }$

let's use a serial expansion

√(1 ±x) = 1 ± ½ x +…

we substitute

vₐ = v_ c ( $(1 + \frac{1}{2} \frac{\Delta M}{M} ) \ ( 1 - \frac{1}{2} \frac{\Delta R}{R} )$ )

we make the product and keep the terms linear

vₐ = v_c $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$

5 0

2 years ago

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

(c) $8.66in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):

$c^2=a^2+b^2-2abcos(\gamma)$

$c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in$

4 0

3 years ago

A 10 Kg ball is rolling at 2.5 m/s. It is then hit from behind with a bat that puts a 300 N force on the ball for a quick .3 sec

vazorg [7]

Answer: g. gg g rfrcdv

Explanation:

vfv g bygyb

7 0

3 years ago